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3 years ago

12

A fountain is made up of two semicircles and a quarter circle. Find the perimeter of the fountain. Round to the nearest hundredt

h.

1 answer:

brilliants [131]3 years ago

8 0

So that mean 12x12=144quarter cm

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A student takes an exam containing 18 true or false questions. If the student guesses, what is the probability that he will get

V125BC [204]

Use this formula.
and solution is:［（18！）/（6！×（18-6）！）］×（1/2）^6×（1-1/2）^（18-6）≈0.0781604
7.8%

3 0

2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

Which of the following best describes the equation below?

Reika [66]

4(x + 9) + 5x = 9x - 36

=> 4x + 36 + 5x = 9x - 36

=> 9x + 36 = 9x - 36

=> 9x - 9x = -36 - 36

=> 0 = -72

You can see that the equation doesn't hold true.

So there is no solution for this equation.

Answer is C. no real solutions

5 0

3 years ago

An arithmetic sequence with a third term of 8 and a constant difference of 5

Vanyuwa [196]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ \hrulefill\\[0.5em] a_3=8\\ n=3\\ d=5 \end{cases} \\\\\\ a_3=a_1+(3-1)5\implies 8=a_1+(2)5 \\\\\\ 8=a_1+10\implies -2=a_1 \\\\\\ \begin{cases} a_1=-2\\ d=5 \end{cases}\implies a_n=-2+(n-1)d$

5 0

3 years ago

What is the value of f(2) for f(x)=2x5−8x4+3x3−4x2+9x−1?(1 point)

Yanka [14]

Answer:

-39

Step-by-step explanation:

just took the quick check

4 0

2 years ago